$ D = \left[\begin{array}{rrr}1 & -2 & 2 \\ 3 & 1 & -1\end{array}\right]$ $ E = \left[\begin{array}{rr}2 & 1 \\ -2 & 3 \\ 5 & 4\end{array}\right]$ What is $ D E$ ?
Answer: Because $ D$ has dimensions $(2\times3)$ and $ E$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D E = \left[\begin{array}{rrr}{1} & {-2} & {2} \\ {3} & {1} & {-1}\end{array}\right] \left[\begin{array}{rr}{2} & \color{#DF0030}{1} \\ {-2} & \color{#DF0030}{3} \\ {5} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{2}+{-2}\cdot{-2}+{2}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{2}+{-2}\cdot{-2}+{2}\cdot{5} & ? \\ {3}\cdot{2}+{1}\cdot{-2}+{-1}\cdot{5} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{2}+{-2}\cdot{-2}+{2}\cdot{5} & {1}\cdot\color{#DF0030}{1}+{-2}\cdot\color{#DF0030}{3}+{2}\cdot\color{#DF0030}{4} \\ {3}\cdot{2}+{1}\cdot{-2}+{-1}\cdot{5} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{2}+{-2}\cdot{-2}+{2}\cdot{5} & {1}\cdot\color{#DF0030}{1}+{-2}\cdot\color{#DF0030}{3}+{2}\cdot\color{#DF0030}{4} \\ {3}\cdot{2}+{1}\cdot{-2}+{-1}\cdot{5} & {3}\cdot\color{#DF0030}{1}+{1}\cdot\color{#DF0030}{3}+{-1}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}16 & 3 \\ -1 & 2\end{array}\right] $